A man pushes a box along a flat, frictionless surface using a force of 500 N. The box was moved a distance of 2.5 m. The actual work done by the worker was

The actual work done by the worker is 1250 J.

Work can be defined as the product of force and distance. The s.i unit of work is Joules

The formula of work is

W = Fd Equation 1

Where W = work done by the worker, F = Force, d = distance.

From the question,

Given: F = 500 N, d =  2.5 m

Substitute these values into equation 1

W = 500×2.5

W = 1250 J

Hence, the actual work done by the worker is 1250 J

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Workdone = 1250Nm

Explanation:

Given the following data;

Force, F = 500N

Distance, d = 2.5m

Workdone is given by the formula;

Substituting into the equation, we have

Workdone = 1250Nm

Therefore, the actual work done by the worker is 1250 Newton-meter.

a charge 3q is at the origin, and a charge −2q is on the positive x axis at x=a. part a where on the x-axis would you place a third charge so it would experience no net electric force? enter an expression for the exact answer, which will involve a square-root sign. (do not enter an approximate decimal answer.)